I was confused about the above decomoposition: $$A_{+}-A_{-}=(A-I)(\chi_A(1,\|A\|)+\chi_A(0,1)),$$ but $\chi_A(1,\|A\|)+\chi_A(0,1)$ may not be equal to $\chi_A(0,\|A\|)$, which implies that we cannot have $A_{+}-A_{-}=A-I$.
Can anyone point out the key point. THANKS!
$A-R_A$ is not $A-I$. It's $(A-I)\,1_{(0,\|A\|]\setminus\{1\}}(A)$.
It's a straightforward computation. Since it is functional calculus, it is enough to show the equality for functions, no operators involved.
Let $c=\|A\|$. Note that $1_{\{1\}}(t)=t\,1_{\{1\}}(t)$. You have, working on $[0,c]$, \begin{align} (t-1)\,1_{(1,c]}(t)-(1-t)\,1_{(0,1)}(t)+1_{(0,c]}(t) &=(t-1)\,1_{(0,c]\setminus\{1\}}(t)+1_{(0,c]}(t)\\[0.3cm] &=t\,1_{(0,c]\setminus\{1\}}(t)-1_{(0,c]\setminus\{1\}}(t)+1_{(0,c]}(t)\\[0.3cm] &=t\,1_{(0,c]\setminus\{1\}}(t)+1_{\{1\}}(t)\\[0.3cm] &=t\,1_{(0,c]\setminus\{1\}}(t)+t\,1_{\{1\}}(t)\\[0.3cm] &=t\,1_{(0,c]}(t)\\[0.3cm] &=t. \end{align}